(a) Differentiate \(y = (3 - 2x)^3 + 24x\) with respect to \(x\):

\(\frac{dy}{dx} = 3(3 - 2x)^2(-2) + 24 = -6(3 - 2x)^2 + 24\).

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -12(3 - 2x)(-2) = 24(3 - 2x)\).

(b) Set \(\frac{dy}{dx} = 0\):

\(-6(3 - 2x)^2 + 24 = 0\)

\(6(3 - 2x)^2 = 24\)

\((3 - 2x)^2 = 4\)

\(3 - 2x = \pm 2\)

\(x = \frac{1}{2}\) or \(x = \frac{5}{2}\)

For \(x = \frac{1}{2}\), \(y = (3 - 2 \times \frac{1}{2})^3 + 24 \times \frac{1}{2} = 20\)

For \(x = \frac{5}{2}\), \(y = (3 - 2 \times \frac{5}{2})^3 + 24 \times \frac{5}{2} = 52\)

(c) Evaluate \(\frac{d^2y}{dx^2}\) at the stationary points:

For \(x = \frac{1}{2}\), \(\frac{d^2y}{dx^2} = 24(3 - 2 \times \frac{1}{2}) = 48\), indicating a minimum.

For \(x = \frac{5}{2}\), \(\frac{d^2y}{dx^2} = 24(3 - 2 \times \frac{5}{2}) = -48\), indicating a maximum.